Supermarket

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17634		Accepted: 7920

題目連結：http://poj.org/problem?id=1456

Description:

A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σ_x∈Sell

px. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

Input:

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output:

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Input:

4  50 2  10 1   20 2   30 1

7  20 1   2 1   10 3  100 2   8 2
   5 20  50 10

Sample Output:

80
185

題意：

給出n個商品以及它們的利潤和截至日期，每天只能賣出一個商品，截止日期過後就不能出售了，問最大利潤是多少。

 

題解：

第一反應肯定是採用貪心的策略，選擇一個最大利潤的出售，但是因為這裡有個日期，這裡的貪心不一定可以得到最後的最大利潤。

所以我們也應該考慮日期，發現當出售商品在其截至日期時，是最優的，因為這將不會影響之前商品的出售。

這裡的貪心我們可以採用優先佇列對其進行優化，並且日期要倒過來列舉（結合貪心策略想想）。

這裡我用的是並查集對其進行優化，商品出售的時間儘可能得接近其截至日期，如果在那天商品出售了，又有另外一個商品在那天出售，那麼我們就挪到前一天進行出售。

我們就用並查集維護離截止日期最近的空閒時間，查詢速度挺快的。

 

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N = 10005; 
int f[N];
int n;

struct node{
    int p,d;
    bool operator < (const node &A)const{
        return A.p<p;
    }
}a[N];
int find(int x){return f[x]==x ? f[x] : f[x]=find(f[x]);}
int main(){
    while(~scanf("%d",&n)){
        for(int i=1;i<=10000;i++) f[i]=i;
        for(int i=1;i<=n;i++) scanf("%d%d",&a[i].p,&a[i].d);
        sort(a+1,a+n+1);
        int ans = 0;
        for(int i=1;i<=n;i++){
            int fd = find(a[i].d);
            if(fd>0){//存在空閒時間
                ans+=a[i].p;
                f[fd]=fd-1;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}